AC to DC conversion using single supply opamp

Thread Starter

Mayuri1234

Joined Apr 7, 2014
7
hello,

I have a single supply(3.3v) available in my circuit. I have an ac input current coming in the circuit ranging from 0 to 350 mA. I have a burden resistor of 10ohms,2 Watt connected across the ac input. Across burden resistor ,I get an equivalent vrms. I need an ac to dc conversion circuit to convert this vrms into dc voltage using single supply opamp. Please do not suggest a diode. I have designed a circuit using opamp and a bulk capacitor but it is taking time to smoothen the dc voltage. The ac input current coming to the circuit has to be detected within 100ms but this circuit needs the input to stay for more than 200 ms. can you suugest some alternative circuit solution.circuit.png
 

#12

Joined Nov 30, 2010
18,076
It seems obvious to me that you're talking about a Precision Rectifier. One op-amp, analog speed, no voltage offset, and you can scale the peak voltage down to its equivalent in RMS with two resistors.
 

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crutschow

Joined Mar 14, 2008
23,335
It seems obvious to me that you're talking about a Precision Rectifier. One op-amp, analog speed, no voltage offset, and you can scale the peak voltage down to its equivalent in RMS with two resistors.
#12, sorry but that circuit doesn't work, as I discovered when I tried to simulate it. :oops:
Using that circuit with a non-inverting configuration allows the negative voltage to come through the diode and the resistors to the output.
In effect you get the sinewave out with a truncated negative half-cycle.

Below is the LTspice simulation of a single-supply, full-wave, precision rectifier that uses no diodes.
It does require rail-rail op amps.

upload_2017-3-2_13-49-10.png
 
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Alec_t

Joined Sep 17, 2013
10,361
The voltage developed across a 10Ω resistor exceeds your supply voltage!
 
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#12

Joined Nov 30, 2010
18,076
I didn't put the output capacitor on it. I assumed people who could design the circuit in post #1 would only need the concept to figure out it needs a capacitor, the feedback resistors should be deleted, and the negative half cycle must be ignored because a single supply op-amp can not respond to the negative half cycle.

Cutting the sine peak to its equivalent in RMS will get your 3.5Vpeak down to 2.479V which is inside your power supply range of 3.3VDC.

The input being a sine wave means measuring one peak voltage will properly represent the true amplitude of the whole sine wave.
 

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Thread Starter

Mayuri1234

Joined Apr 7, 2014
7
#12, sorry but that circuit doesn't work, as I discovered when I tried to simulate it. :oops:
Using that circuit with a non-inverting configuration allows the negative voltage to come through the diode and the resistors to the output.
In effect you get the sinewave out with a truncated negative half-cycle.

Below is the LTspice simulation of a single-supply, full-wave, precision rectifier that uses no diodes.
It does require rail-rail op amps.

View attachment 121637
This is good. How should I get a smooth DC ?
 

EM Fields

Joined Jun 8, 2016
583
hello,

I have a single supply(3.3v) available in my circuit. I have an ac input current coming in the circuit ranging from 0 to 350 mA. I have a burden resistor of 10ohms,2 Watt connected across the ac input. Across burden resistor ,I get an equivalent vrms. I need an ac to dc conversion circuit to convert this vrms into dc voltage using single supply opamp. Please do not suggest a diode. I have designed a circuit using opamp and a bulk capacitor but it is taking time to smoothen the dc voltage. The ac input current coming to the circuit has to be detected within 100ms but this circuit needs the input to stay for more than 200 ms. can you suugest some alternative circuit solution
Yes.

Instead of doing the rectifier-filter thing, consider using a peak-hold circuit like this:
Peak hold concept.png
Since the input frequency is 50Hz, the time from any zero-crossing to the next peak will be 5 milliseconds.
Therefore, detect a positive-going zero-crossing and generate a delay which, when it decays, will generate a sample pulse which will be centered 5 milliseconds away from the zero crossing, which is where the positive peak of the AC signal will be. Use that sample pulse to momentarily connect the AC signal to a capacitor and charge it up to the ac signal's peak.

The sample and hold part of the circuit is flaky, but it works well enough with a 350mA AC current source to simulate and illustrate the concept in hardware.

In reality, everything is happening so slowly that most (if not all) of the thing could be easily implemented in software.
 
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